Symmetry of Lorenz System

Question

This is one idea I haven't quite understood how to apply.

The Lorenz system:

$\dot{x}=\sigma\left ( y-x \right )$

$\dot{y}=rx-y-xz$

$\dot{z}=-bz+xy$

I have worked out all the fixed points.

I would like to show that the system is reversible under

$x \mapsto -x$ and $y \mapsto -y$

I.e, system is symmetric under an inversion through the z-axis.

Should I be substituting all the above RHS equation by $x \mapsto -x$ and $y \mapsto -y$? I haven't read through Strogaz text and he handwaved quite a fair bit without providing the meat of the technicality.

Any explanation would be greatly appreciated.

Answer 1

When $x \mapsto -x , y \mapsto -y $ then

$-\dot{x} = \sigma((-y) - (-x))$

$-\dot{y} = r(-x) - (-y) - (-x)(z)$

$\dot{z} = -bz + (-x)(-y)$

Which are the same Lorenz equations.

Also if you want a more reference,a matrix based approach you can refer http://planetmath.org/naturalsymmetryofthelorenzequation

Enjoy learning Lorenz equations :)

Answer 2

Let us set $x = -X$, $y = -Y$, and $z = Z$. Thus, $\dot{x} = -\dot{X}$, $\dot{y} = -\dot{Y}$, and $\dot{z} = \dot{Z}$. The Lorenz system in terms of $\lbrace x, y , z\rbrace$ rewrites as $$ \left\lbrace \begin{aligned} & \dot{X} = \sigma \left( Y - X \right) \\ & \dot{Y} = rX - Y - XZ \\ & \dot{Z} = -bZ + XY \end{aligned} \right. $$ in terms of $\lbrace X, Y , Z\rbrace$, which is the same differential system as before.